Question :
For the remaining questions, consider a 4-bit block cipher, described in hexadecimal by the following table:
|
Plaintext
|
Ciphertext
|
Plaintext
|
Ciphertext
|
|
0
|
a
|
8
|
e
|
|
1
|
c
|
9
|
d
|
|
2
|
f
|
a
|
0
|
|
3
|
6
|
b
|
7
|
|
4
|
3
|
c
|
5
|
|
5
|
8
|
d
|
b
|
|
6
|
4
|
e
|
9
|
|
7
|
2
|
f
|
1
|
You can think of this as a simple substitution cipher for hexadecimal digits. There is no "key" other than the table itself.
For this question, you will perform encryption and decryption using the same cipher described above, but in CBC mode. Recall that you can convert hexadecimal digits to binary to do the XOR operation.
Encrypt the 4-block message face using the above block cipher in CBC mode with an IV of 6 (0110 in binary).
Decrypt the message 2d21 using the above cipher in CBC mode with an IV of 6